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Rom4ik [11]
3 years ago
7

The nearest star to Earth is Proxima Centauri, 4.3 light-years away.

Physics
1 answer:
PIT_PIT [208]3 years ago
7 0

Answer:

a) v = 2,152 10⁸ m / s   b)   t = 2.71 10⁸ s  or t =  85.93 year

Explanation:

a) In this special relativity exercise we have that time is measured in the same ship, so it is the proper time,

    v = d / t

Let's reduce the distance to the SI system

    d = 4.3 l and (9.46 1015 m / 1ly) = 40.678 10¹⁵ m

    t = 5.0 y (365 day / 1 y) (24 h / 1 day) (3600s / 1h) = 1.89 10⁸ s

Let's calculate

    v = 40.678 10¹⁵ / 1.89 10⁸

    v = 2,152 10⁸ m / s

b) The time seen from the ground for which the ship moves is given by

     t = t₀ / √ (1- (v/c)²)

Let's calculate

     t = 1.89 10⁸ / √ (1 - (2.152 / 2.998)²)

     t = 1.89 10⁸ / 0.6962

     t = 2.71 10⁸ s

Let's reduce this time to years

    t = 2.71 10⁸ s (1h / 3600s) (1day / 24h) (1 and / 365 d)

    t =  85.93 year

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4 0
3 years ago
A swimmer swims at 5 m/s. How long would it take to swim 5 laps of a 50m pool?​
Gre4nikov [31]
About 5 hours gooood luck
3 0
1 year ago
A distant planet with a mass of (7.2000x10^26) has a moon with a mass of (5.0000x10^23). The distance between the planet and the
BARSIC [14]

Answer:

Explanation:

This is a simple gravitational force problem using the equation:

F_g=\frac{Gm_1m_2}{r^2} where F is the gravitational force, G is the universal gravitational constant, the m's are the masses of the2 objects, and r is the distance between the centers of the masses. I am going to state G to 3 sig fig's so that is the number of sig fig's we will have in our answer. If we are solving for the gravitational force, we can fill in everything else where it goes. Keep in mind that I am NOT rounding until the very end, even when I show some simplification before the final answer.

Filling in:

F_g=\frac{(6.67*19^{-11})(7.2000*10^{26})(5.0000*10^{23})}{(6.10*10^{11})^2} I'm going to do the math on the top and then on the bottom and divide at the end.

F_g=\frac{2.4012*10^{40}}{3.721*10^{23}} and now when I divide I will express my answer to the correct number of sig dig's:

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8 0
3 years ago
If an object is projected upward from ground level with an initial velocity of 144144 ft per​ sec, then its height in feet after
Liula [17]

Answer:

4.5 s, 324 ft

Explanation:

The object is projected upward with an initial velocity of

v_0 = 144 ft/s

The equation that describes its height at time t is

s(t) = -16t^2 + 144 t (1)

where t, the time, is measured in seconds.

In order to find the time it takes for the object to reach the maximum height, we must find an expression for its velocity at time t, which can be found by calculating the derivative of the position, s(t):

v(t) = s'(t) = -32t +144 (2)

At the maximum heigth, the vertical velocity is zero:

v(t) = 0

Substituting into the equation above, we find the corresponding time at which the object reaches the maximum height:

0=-32t+144\\t=\frac{144}{32}=4.5 s

And by substituting this value into eq.(1), we also find the maximum height:

s(t) = -16(4.5)^2+144(4.5)=324 ft

3 0
3 years ago
A train is travelling towards the station on a straight track. It is a certain distance from the station when the engineer appli
AleksandrR [38]

Answer:

500 m

Explanation:

t = Time taken

u = Initial velocity = 50 m/s

v = Final velocity = 0

s = Displacement

a = Acceleration = -2.5 m/s²

Equation of motion

v=u+at\\\Rightarrow 0=50-2.5\times t\\\Rightarrow \frac{-50}{-2.5}=t\\\Rightarrow t=20\ s

Time taken by the train to stop is 20 seconds

s=ut+\frac{1}{2}at^2\\\Rightarrow s=50\times 20+\frac{1}{2}\times -2.5\times 20^2\\\Rightarrow s=500\ m

∴ The engineer applied the brakes 500 m from the station

4 0
3 years ago
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